# We denote by σ(n) the sum of the divisors of the natural number n

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Problem 10711

(American Mathematical Monthly, Vol.106, February 1999) Proposed by F. Luca (Germany).

A natural number isperfect if it is the sum of its proper divisors. Prove that two consecutive num- bers cannot both be perfect.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Firenze, viale Morgagni 67/A, 50134 Firenze, Italy.

We denote by σ(n) the sum of the divisors of the natural number n. It is easy to see that the arithmetic function σ is multiplicative that is

σ(a · b) = σ(a) · σ(b) iff (a, b) = 1.

Then n is a perfect number if and only if σ(n) = 2n. Moreover the following two elementary properties of perfect numbers hold:

(i) if n is even then n is perfect if and only if

n= 2q−1(2q−1) where q and 2q−1 are primes.

(ii) if n is an odd perfect number then it must have the form n= p4m+1s2 where p is a prime such that [p]4= 1 and (p, s) = 1.

For a proof of the above results see for example L. Dickson, History of the theory of numbers, vol.

1, 1919, pp. 1-33. Note that the question as to whether there exist odd perfect numbers is as yet unsolved.

Therefore, by (i), two consecutive numbers are both perfect if

n= 2q−1(2q−1) − 1 or n+= 2q−1(2q−1) + 1

is perfect. The prime q is not 2 because neither n= 5 nor n+= 7 is perfect. Hence q = 2a + 1 ≥ 3.

1) Assume that n is perfect. Then

[n]4= [4a]4·[2 · 4a−1]4+ [−1]4= 4.

By (ii), n = p4m+1s2 with [p]4 = 1. Since s is odd then either [s]4 = 1 or [s]4 = 3 and in both cases [s2]4= 1. So we have the contradiction

4= [n]4= [p]44m+1·[s2]4= 4. 2) Assume that n+ is perfect. Then

[n+]3= [4a]3·[2 · 4a−1]3+ 3= 3+ 3= 3.

By (ii), n+= p4m+1s2 where p is a prime such that (p, s) = 1. Now, it easy to verify that [p4m+1]3·[s2]3= 3 iff [p]3= 3 and [s2]3= 1.

Moreover, since p is prime,

[σ(p4m+1)]3= [1 + p + . . . + p4m+ p4m+1]3= [1 + 2 + . . . + 1 + 2]3= 3. By the multiplicative property of σ, we have the contradiction

3= [2n+]3= [σ(n+)]3= [σ(p4m+1)]3·[σ(s2)]3= 3.

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